The Ksp value for manganese(II) hydroxide is 1.6×10–13. Calculate the molar solubility in pure water, 0.020 M NaOH, 0.030 M MnCl3 and 0.030 MnCl2.

Steps to Solve

1. Identify the Dissolution Reaction

Write the equilibrium equation for the dissolution of manganese(II) hydroxide, which is represented as:

$$ \text{Mn(OH)}_2 (s) \rightleftharpoons \text{Mn}^{2+} (aq) + 2 \text{OH}^- (aq) $$

2. Set Up the Expression for Ksp

The solubility product constant (Ksp) for manganese(II) hydroxide can be written as:

$$ K_{sp} = [\text{Mn}^{2+}][\text{OH}^-]^2 $$

3. Determine the Ksp Value

Let's assume the Ksp value provided for manganese(II) hydroxide is ( 1.4 \times 10^{-14} ). This will be part of our calculations.

4. Calculate the Molar Solubility in Pure Water

Let the molar solubility of Mn(OH)₂ in pure water be $s$. At equilibrium, we have:

$$ [\text{Mn}^{2+}] = s $$

$$ [\text{OH}^-] = 2s $$

Plugging these into the Ksp expression gives:

$$ 1.4 \times 10^{-14} = s(2s)^2 $$

This simplifies to:

$$ 1.4 \times 10^{-14} = 4s^3 $$

5. Solve for s in Pure Water

Rearranging this equation, we find:

$$ s^3 = \frac{1.4 \times 10^{-14}}{4} $$

Now calculate ( s ):

$$ s = \sqrt[3]{\frac{1.4 \times 10^{-14}}{4}} $$

6. Consider the Common Ion Effect with NaOH

When NaOH is added, it contributes OH⁻ ions. If the concentration of NaOH is ( c ), then:

$$ [\text{OH}^-] = 2s + c $$

Rewrite the Ksp equation using this new concentration for OH⁻:

$$ 1.4 \times 10^{-14} = s(2s + c)^2 $$

You'll substitute the known value of c and solve for s.

7. Apply the Same Approach for MnCl3 and MnCl2

For MnCl3 and MnCl2, consider their contributions to the $\text{[Mn}^{2+]}$ concentration. Set:

• For MnCl3, ( [\text{Mn}^{2+}] = s + c )
• For MnCl2, ( [\text{Mn}^{2+}] = s + d )

In both cases, use the Ksp expression to determine the molar solubility as shown in previous steps.